Asked by jj reddick
                the following data were collected when titrating 25.00 mL samples of HCL with 0.511 M NaOH:
initial burette reading (mL): trial 1=1.06 trial2=0.04 trial3=1.03
final burette reading (mL): trial1= 27.60 trial2=26.21 trial3=27.22
            
            
        initial burette reading (mL): trial 1=1.06 trial2=0.04 trial3=1.03
final burette reading (mL): trial1= 27.60 trial2=26.21 trial3=27.22
Answers
                    Answered by
            bobpursley
            
    MolarityAcid*.025=.511M*volumebase
averaging the volume of base:
27.60-1.06=26.54
26.21-.04=26.17
27.22-1.03=26.18
Now I am wondering what your question is.
    
averaging the volume of base:
27.60-1.06=26.54
26.21-.04=26.17
27.22-1.03=26.18
Now I am wondering what your question is.
                    Answered by
            jj reddick
            
    using this data what is the [HCL]?
    
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